Permutation and combination calculator
Does order matter? That one question decides between a permutation and a combination. Small cases are listed so you can count them yourself.
nPr = 5! / (5 − 3)! · nCr = 5! / (3! · 2!)
All 60 arrangements of A, B, C, D, E
ABC ABD ABE ACB ACD ACE ADB ADC ADE AEB AEC AED BAC BAD BAE BCA BCD BCE BDA BDC BDE BEA BEC BED CAB CAD CAE CBA CBD CBE CDA CDB CDE CEA CEB CED DAB DAC DAE DBA DBC DBE DCA DCB DCE DEA DEB DEC EAB EAC EAD EBA EBC EBD ECA ECB ECD EDA EDB EDC
The 10 groups (order ignored)
ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
Each group of 3 letters appears as 6 different arrangements in the first list (3! = 6).
Order or no order
Choosing a president, secretary and treasurer from 5 people is a permutation: Ana-Ben-Cy is different from Cy-Ben-Ana. Choosing a 3-person committee is a combination: those are the same three people. For every group of r items there are r! orderings, which is exactly the factor between the two counts: nPr = nCr × r!.
nPr = n! / (n − r)! nCr = n! / (r!(n − r)!)
With repetition
- Permutations with repetition, like a 4-digit PIN: 10⁴ = 10,000.
- Combinations with repetition, like choosing 3 scoops from 5 flavours where repeats are allowed: C(5 + 3 − 1, 3) = 35 (the "stars and bars" method).
Where these show up
nCr is the C(n, k) in the binomial formula, it counts lottery tickets (C(49, 6) = 13,983,816), and it counts the pairs in the birthday paradox (C(23, 2) = 253).
Still not clicking? Ask the tutor
An AI tutor that sees the numbers you entered above and explains them. It can make mistakes, so check anything important against the worked steps.
Questions students ask
What is the difference between a permutation and a combination?
In a permutation order matters (ABC and CBA are different); in a combination it does not (they are the same group). There are always k! times as many permutations as combinations.
What are the formulas?
nPr = n! / (n − r)! and nCr = n! / (r!(n − r)!). With repetition allowed, permutations are n^r and combinations are C(n + r − 1, r).
How many lottery combinations are there for 6 from 49?
C(49, 6) = 13,983,816, so one ticket has about a 1 in 14 million chance of the jackpot.
What is 0!?
0! = 1. There is exactly one way to arrange nothing, and this definition makes the formulas work when r = n or r = 0.